B. Rational Functions. If $R(x) = \frac{P(x)}{Q(x)},$ where $$P$$ and $$Q$$ are polynomials, it follows at once from § 113, (5) that $R'(x) = \frac{P'(x)Q(x) – P(x)Q'(x)}{\{Q(x)\}^{2}},$ and this formula enables us to write down the derivative of any rational function. The form in which we obtain it, however, may or may not be the simplest possible. It will be the simplest possible if $$Q(x)$$ and $$Q'(x)$$ have no common factor, i.e. if $$Q(x)$$ has no repeated factor. But if $$Q(x)$$ has a repeated factor then the expression which we obtain for $$R'(x)$$ will be capable of further reduction.

It is very often convenient, in differentiating a rational function, to employ the method of partial fractions. We shall suppose that $$Q(x)$$, as in § 116, is expressed in the form $a_{0}(x – \alpha_{1})^{m_{1}} (x – \alpha_{2})^{m_{2}}\dots (x – \alpha_{\nu})^{m_{\nu}}.$ Then it is proved in treatises on Algebra1 that $$R(x)$$ can be expressed in the form \begin{aligned} \Pi(x) &+ \frac{A_{1, 1}}{x – \alpha_{1}} + \frac{A_{1, 2}}{(x – \alpha_{1})^{2}} + \dots + \frac{A_{1, m_{1}}}{(x – \alpha_{1})^{m_{1}}}\\ &+ \frac{A_{2, 1}}{x – \alpha_{2}} + \frac{A_{2, 2}}{(x – \alpha_{2})^{2}} + \dots + \frac{A_{2, m_{2}}}{(x – \alpha_{2})^{m_{2}}} + \dots,\end{aligned} where $$\Pi(x)$$ is a polynomial; i.e. as the sum of a polynomial and the sum of a number of terms of the type $\frac{A}{(x – \alpha)^{p}},$ where $$\alpha$$ is a root of $$Q(x) = 0$$. We know already how to find the derivative of the polynomial: and it follows at once from Theorem (4) of § 113, or, if $$\alpha$$ is complex, from its extension indicated in § 114, that the derivative of the rational function last written is $-\frac{pA(x -\alpha)^{p-1}}{(x – \alpha)^{2p}} = -\frac{pA}{(x – \alpha)^{p+1}}.$

We are now able to write down the derivative of the general rational function $$R(x)$$, in the form $\Pi'(x) – \frac{A_{1, 1}}{(x – \alpha_{1})^{2}} – \frac{2A_{1, 2}}{(x – \alpha_{1})^{3}} – \dots – \frac{A_{2, 1}}{(x – \alpha_{2})^{2}} – \frac{2A_{2, 2}}{(x – \alpha_{2})^{3}} – \dots.$ Incidentally we have proved that the derivative of $$x^{m}$$ is $$mx^{m-1}$$, for all integral values of $$m$$ positive or negative.

The method explained in this section is particularly useful when we have to differentiate a rational function several times (see Exs. XLV).

Example XLII

1. Prove that $\frac{d}{dx}\left(\frac{x}{1 + x^{2}}\right) = \frac{1 – x^{2}}{(1 + x^{2})^{2}},\quad \frac{d}{dx}\left(\frac{1 – x^{2}}{1 + x^{2}}\right) = -\frac{4x}{(1 + x^{2})^{2}}.$

2. Prove that $\frac{d}{dx}\left(\frac{ax^{2} + 2bx + c}{Ax^{2} + 2Bx + C}\right) = \frac{(ax + b) (Bx + C) – (bx + c) (Ax + B)}{(Ax^{2} + 2Bx + C)^{2}}.$

3. If $$Q$$ has a factor $$(x – \alpha)^{m}$$ then the denominator of $$R’$$ (when $$R’$$ is reduced to its lowest terms) is divisible by $$(x – \alpha)^{m+1}$$ but by no higher power of $$x – \alpha$$.

4. In no case can the denominator of $$R’$$ have a simple factor $$x – \alpha$$. Hence no rational function (such as $$1/x$$) whose denominator contains any simple factor can be the derivative of another rational function.

1. See, e.g., Chrystal’s Algebra, vol. i, pp. 151 et seq.↩︎